Codeforces 85D. Sum of Medians

题目大意：

对一个集合\(A = \emptyset\)，有插入元素、删除元素、求“和”三种操作。设某个时刻集合A为：

\[A= {a_1, a_2, a_3, \ldots,a_k} \]

定义其“和”为：

\[\sum_{i\ mod\ 5\ ==\ 3}^{i\ \leq\ k}a_i \]

对每次求“和”操作，输出其“和”

英文原题

D. Sum of Medians

time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is called the middle element of a sorted array (it's the third largest element for a group of five). To increase the algorithm's performance speed on a modern video card, you should be able to find a sum of medians in each five of the array.

A sum of medians of a sorted k-element set S = {a1, a2, ..., ak}, where a1 < a2 < a3 < ... < ak, will be understood by as

The operator stands for taking the remainder, that is stands for the remainder of dividing x by y.

To organize exercise testing quickly calculating the sum of medians for a changing set was needed.
Input

The first line contains number n (1 ≤ n ≤ 105), the number of operations performed.

Then each of n lines contains the description of one of the three operations:

add x — add the element x to the set;
del x — delete the element x from the set;
sum — find the sum of medians of the set. 

For any add x operation it is true that the element x is not included in the set directly before the operation.

For any del x operation it is true that the element x is included in the set directly before the operation.

All the numbers in the input are positive integers, not exceeding 109.
Output

For each operation sum print on the single line the sum of medians of the current set. If the set is empty, print 0.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams (also you may use the %I64d specificator).

Examples

Input

Copy

6
add 4
add 5
add 1
add 2
add 3
sum

Output

3

Input

Copy

14
add 1
add 7
add 2
add 5
sum
add 6
add 8
add 9
add 3
add 4
add 10
sum
del 1
sum

Output

5
11
13

题解

可以发现某个元素是否被计入和中与其当前的序号有关，于是把所有数离线，离散化。
虽然平衡树也可做，但是本题有特殊性质，插入元素的排序是固定的，考虑权值线段树能否胜任
以其离散化后的顺序为节点，每个节点维护区间序列中\(mod\ 5 = 0/1/2/3/4\)的数的原数和
合并的时候，左区间直接加入，右区间需要左区间的大小对右区间的和整体位移。
因此每个节点还要维护数的个数

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
inline long long max(long long a, long long b){return a > b ? a : b;}
inline long long min(long long a, long long b){return a < b ? a : b;}
inline void swap(long long &x, long long &y){long long tmp = x;x = y;y = tmp;}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long MAXN = 100000 + 10;
struct Node
{
	long long num[5], size;
	Node(){num[0] = num[1] = num[2] = num[3] = num[4] = size = 0;}
}node[MAXN << 2];
void pushup(long long o)
{
	long long l = o << 1, r = o << 1 | 1, size = node[l].size;
	for(long long i = 0;i <= 4;++ i)
		node[o].num[i] = node[l].num[i];
	for(long long i = 0;i <= 4;++ i)
		node[o].num[(size + i) % 5] += node[r].num[i];
	node[o].size = size + node[r].size
;}
long long q; 
long long cnt[MAXN], numm[MAXN], tot, ma;
long long num[MAXN], pos[MAXN], tp[MAXN];

//把k这个元素插入 / 删除 
void modify(long long k, long long o = 1, long long l = 1, long long r = ma)
{
	if(l == r && l == abs(k))
	{
		if(k > 0) node[o].num[1] += pos[k], ++ node[o].size; 
		else node[o].num[1] -= pos[-k], -- node[o].size;
		return;
	}
	long long mid = (l + r) >> 1;
	if(mid >= abs(k)) modify(k, o << 1, l, mid);
	else modify(k, o << 1 | 1, mid + 1, r);
	pushup(o);
}
char s[10];
bool cmp(long long a, long long b)
{
	return numm[a] < numm[b];
}
int main()
{
	read(q);
	for(long long i = 1;i <= q;++ i)
	{
		scanf("%s", s + 1);
		if(s[1] == 'a')
		{
			read(numm[++ tot]), cnt[tot] = tot;
			tp[i] = 1;
		}
		else if(s[1] == 'd')
		{
			read(numm[++ tot]), cnt[tot] = tot;
			tp[i] = -1;
		}
	}
	std::sort(cnt + 1, cnt + 1 + tot, cmp);
	for(long long i = 1;i <= tot;++ i)
	{
		if(i == 1 || numm[cnt[i]] != numm[cnt[i - 1]]) ++ ma, pos[ma] = numm[cnt[i]];
		num[cnt[i]] = ma;
	}
	for(long long i = 1, k = 1;i <= q;++ i)
	{
		if(tp[i]) modify(tp[i] * num[k]), ++ k;
		else printf("%I64d\n", node[1].num[3]);
	}
    return 0;
}